How do you integrate tan^5(x)?

[tangibleLime]

Integrating tan^5(x)... [UNSOLVED]

Homework Statement

\int tan^5(x) dx

Homework Equations

tan^2(x) = sec^2(x) - 1

The Attempt at a Solution

First I split it up into (3+2).
\int tan^3(x)tan^2(x) dx

Using the identity for tangent squared...
\int tan^3(x)(sec^2(x) - 1) dx

Distributing the tan cubed...
\int tan^3(x)sec^2(x)-tan^3(x) dx

Braking it into two integrals...
\int tan^3(x)sec^2(x) dx - \int tan^3(x) dx

For the first integral, substituting u = tanx, du = sec^2(x) dx which takes care of the right half of it.
\int u^3 du = \frac{u^4}{4} = \frac{tan^4(x)}{4}

Now for the second integral from several steps above. Breaking it down to take out a tangent to get a tan squared.
\int tan^2(x)tan(x) dx

Using the trig identity for tan squared...
\int (sec^2(x)-1)tan(x) dx

Distribute the tan...
\int sec^2(x)tan(x)-tan(x) dx

Break this into another two integrals...
\int sec^2(x)tan(x) dx - \int tan(x) dx

First off, the right integral is easy, so just do that first...
\int tan(x) dx = ln|sec(x)|

Now substitute u = tanx for the left integral so du = sec^2(x) and take care of that secant squared.
\int u du = \frac{u^2}{2} = \frac{tan^2}{2}

Combining these last two results (the right integral on the first separation way in the beginning)
\frac{1}{2}tan^2(x) - ln|sec(x)|

Combining it with the original left integral and constant for final answer...
\frac{1}{4}tan^4(x) - \frac{1}{2}tan^2(x) - ln|sec(x)| + C

This was found to be incorrect. When I tried to use WolframAlpha to find the derivative to check if it could be an alternate answer, Wolfram couldn't deal with it and instead gave me the natural log of secant. 0_0

Any help would be greatly appreciated.

 

[Fragment]

Nothing wrong with your work. I think you're just forgetting that the integral of tan(x) can be represented as ln|secx| or -log(cos(x)). Try it with cosx and you'll get the same answer as Wolfram.


F

 

[tangibleLime]

Ah, thanks! I had a hunch.